A) 4 : 1
B) 8 : 1
C) 7 : 1
D) 49 : 1
Correct Answer: D
Solution :
From figure \[{{I}_{1}}=\frac{I}{4}\] and \[{{I}_{2}}=\frac{9I}{64}\] Þ \[\frac{{{I}_{2}}}{{{I}_{1}}}=\frac{9}{16}\] By using \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\left( \frac{\sqrt{\frac{{{I}_{2}}}{{{I}_{1}}}}+1}{\sqrt{\frac{{{I}_{2}}}{{{I}_{1}}}}-1} \right)=\left( \frac{\sqrt{\frac{9}{16}}+1}{\sqrt{\frac{9}{16}}-1} \right)=\frac{49}{1}\]You need to login to perform this action.
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